Question #3a121

1 Answer
Sep 25, 2017

Answer:

Here's what I got.

Explanation:

The idea here is that you need to use the name of the compound to write its chemical formula.

As you can see, you're dealing with an ionic compound that contains a transition metal, samarium. The Roman numeral that follows the name of the metal tells you its charge in the compound.

In your case, the #("III")# Roman numeral tells you that the samarium cation carries a #3+# charge in this compound.

Since the bromate anion carries a #1-# charge, you can say that you will need #3# bromate anions to balance the overall positive charge coming from the cation.

This means that the chemical formula for samarium(III) bromate is

#["Sm"]^(3+) + 3["BrO"_3]^(-) -> "Sm"("BrO"_3)_3#

Now, you don't have information about the solubility of samarium(III) bromate in water, so you can assume that you're dealing with a soluble ionic compound.

This means that the dissociation equation will look like this

#"Sm"("BrO"_ 3)_ (3(aq)) -> "Sm"_ ((aq))^(3+) + 3"BrO"_ (3(aq))^(-)#

Consequently, you can say that when #2# moles of samarium(III) bromate dissociate, you get

#2 color(red)(cancel(color(black)("moles Sm"("BrO"_3)_3))) * "3 moles BrO"_3^(-)/(1color(red)(cancel(color(black)("mole Sm"("BrO"_3)_3)))) = color(darkgreen)(ul(color(black)("6 moles BrO"_3^(-))))#

#color(white)(a)#
SIDE NOTE#color(white)(.)# If you assume that samarium(III) bromate is insoluble in water, you can write its dissociation equilibrium as

#"Sm"("BrO"_ 3)_ (3(aq)) rightleftharpoons "Sm"_ ((aq))^(3+) + 3"BrO"_ (3(aq))^(-)#

If you take the fact that two moles of samarium(III) bromate have been dissolved in water to mean that two moles have dissociated in aqueous solution, you can once again say that the solution will contain #6# moles of bromate anions.

The thing to keep in mind here is that depending on the value of the solubility product constant for samarium(III) bromate, you can dissolve #2# moles of this compound, for example, in #"1 L"# of solution or in #"100 L"# of solution.